The Mysterious 8.88 ft-lbs: A Man in Space Story

[bruce2g]

Being aware of the principle of energy conservation, Buddy1 and Buddy2 both are scratching their heads at this point. Buddy1 and Buddy2 both saw -1 energy unit removed from the MCP. Buddy1 wonders: since Buddy2's velocity remained constant, all of the energy applied to the DAP must have come from the MCP so how can the DAP have 4 times the amount of energy? It doesn't seem like 1 energy unit transferred from the MCP will account for the 3 energy units acquired by the DAP. Where did the 2 additional energy units come from? Buddy2 and Mr. FxT are wondering the same thing.

Do the Buddies and Mr. FxT realize that two shots were fired, one to accelerate the DAP, and another to declerate the MCP. Each shot added energy of 1/2 mv^2. The MCP added energy of 1/2 mv^2. It all went to the DAP, to give it energy of 2 mv^2. Where else could the energy from the shots have gone, assuming you believe in conservation of energy?

Bruce

PS: Thanks for giving me the opportunity to repeat the "error" in my previous calculations here; perhaps you or Eyesaw would be kind enough to point it out this time.

 

[Eyesaw]

I analyzed things slightly differently, so maybe my perspective may help a bit. First off, since we are talking about firing a gun (setting off an explosion, in effect) mechanical energy (KE) is not conserved. So that is certainly not an assumption.

So how much energy is released by the firing of the bullet? The bullet gets 1/2mv_1^2, as stated, but momentum conservation tells us that the platform gets pushed back at speed v_2 = (m/M)v_1. So the total KE released is that of the bullet plus platform = 1/2mv_1^2 + 1/2Mv_2^2 = (1 + m/M)(1/2mv_1^2).

Viewed from the "rest" frame, the initial KE of everything before the second firing is 1/2(m + M)v_1^2. After firing, the bullet has speed 2v_1 and the platform has speed v_1 - v_2 = (1 - m/M)v_1, so the final KE of everything is: 1/2m(2v_1)^2 + 1/2M(1 - m/M)^2v_1^2.

Do the arithmetic and you will find that the increase in KE of everything exactly equals the KE released by the firing. There is no missing energy.

The argument made here basically is that measured inside the inertial frame of the second firing, there's a quantity of kinetic energy c, which when viewed by an inertial frame moving at -v, obtains the value c-v. But after taking into account of the -v, the second inertial frame also obtains the value c for the kinetic energy. Now if you only admitted this same logic applies to the speed of light, you'd agree there's no need to warp space-time to conserve the quantity c.

 

[Eyesaw]

Do the Buddies and Mr. FxT realize that two shots were fired, one to accelerate the DAP, and another to declerate the MCP. Each shot added energy of 1/2 mv^2. The MCP added energy of 1/2 mv^2. It all went to the DAP, to give it energy of 2 mv^2. Where else could the energy from the shots have gone, assuming you believe in conservation of energy?

Bruce

PS: Thanks for giving me the opportunity to repeat the "error" in my previous calculations here; perhaps you or Eyesaw would be kind enough to point it out this time.

Well, the way you added up the total kinetic energy in your earlier post was
1/2mv^2 + mv^2 + 1/2mv^2 to get 2mv^2. Since you were viewing from the ground frame, in your example the second explosion actually released 3.0mv^2 of kinetic energy, judging by the final velocity of the left brick of 2v, 1.5mv^2 added to the left brick and -1.5mv^2 added to the right brick. The 1.5mv^2 gave the left brick a final kinetic energy of 2.0mv^2 while the 1.5mv^2 subtracted from the kinetic energy of the right brick (since the right brick was moving at v when the second explosion transferred k.e. against the direction of its motion , i.e. -1.5mv^2) leaving the right brick with a final kinetic energy of -1.0mv^2, viewed from the ground.

You thought the right brick came to rest but it didn't since it has a final kinetic energy of -1.0mv^2.

So a few things you went wrong in your calculation. First, you calculated the wrong amount of total kinetic energy of the system as viewed from the ground. Second, I think it's in error in your example to consider the right brick as having lost kinetic energy to the left brick when it's obvious that the explosion is what transferred kinetic energy to both bricks, just as it was obvious when the explosion happened "at rest" on the ground. A negative value in kinetic energy does not automatically mean a loss in kinetic energy- if that were so, I could probably lose weight just by walking backwards than I normally do.

No, in your case, the 1.50 mv^2 gained by the left brick was generated by the explosive charge, it's the same 0.5mv^2 transferred by the charge in the first explosion when at "rest". The second explosion has more kinetic energy than the first because you are viewing the second explosion from a frame that has v less velocity, which when plugged into the kinetic energy equation, accounts totally for the "extra" energy.

 

[Eyesaw]

Slinkie,

This is very frustrating. It is clear from your last post that you did not read any of my work. (Post # 23, which Eyesore did not quote in its entirety. Please read the original.) If you had read it, you would not think that the fact that the pellet ended up with four times the inital kinetic energy was "absurd". My analysis did not mention momentum conservation at all. It showed quantitatively that the second gun did indeed do three times more work on the pellet than the first, which negates the subsequent reasons that you had for questioning the definition of work. I will invest no more time in a thread in which other participants are unwilling to consider evidence presented to them. Debate is good, but a discussion in which my valid and quantitative points are consistently ignored or obfuscated in order to lend credence to yours, does not constitute an intelligent debate or physical analysis. Please go over my calculations again. Didn't you see how I arrived at the answer that the second gun does 375 J of work, three times more than the first?

As for your question: what is energy? It is an excellent and very deep question. I am not sure when the concept of work as defined in its precise, physical sense first arose, but consider this: there is some quantity in physics having fundamental units kgm²/s² that as we expand physics beyond the realm of mechanics we find is conserved in every conceivable situation, although it has the capability of being transformed into various forms, some more useful for "doing stuff" than others. Examples are mechanical energy (kinetic/potential), thermal energy, EMR, etc. That is all we can say for sure. SO the definition of work as is provides for the simplest setup with which to analyse this amazing conservation law, for if you changed the definition of work to force X time, the resulting change mv caused by this new work would not be dimensionally consistent with any of the other forms of "energy" identified in other realms of physics...and hence would not qualify as energy. I see no benefit to applying the same name to totally different quantities...the accounting books would not balance. In fact, the quantities as defined in your system would not differ from their current roles in physics. They would differ in name only. What you would be calling 'work' would still have the fundamental properties of what we properly call impulse, and nothing would have changed, except that we would be left with a far more contradictory and cumbersome system of terminology in which so called "energy" would refer to at least two different types of quantities, and would therefore be neither transferable nor universally conserved.

Yes, you showed that the second firing did 3 times more work, but did you show why 3 times more work didn't add 3 times more velocity to the pellet? No, the only answer you could give was that because K.E. is defined to be 1/2mv^2. If K.E. was undefined and you started from rest and applied a constant force over 1 distance unit to achieve a velocity v for a particular mass, wouldn't it be reasonable to assume that if you applied the same force over 2 distances, you should achieve a final velocity of 2v? Why isn't this the case? Isn't that a more interesting question than whether you applied the formulas correctly or not?

 

[Doc Al]

The argument made here basically is that measured inside the inertial frame of the second firing, there's a quantity of kinetic energy c, which when viewed by an inertial frame moving at -v, obtains the value c-v. But after taking into account of the -v, the second inertial frame also obtains the value c for the kinetic energy.

You seem to mix up speed with kinetic energy.

Now if you only admitted this same logic applies to the speed of light, you'd agree there's no need to warp space-time to conserve the quantity c.

It is not "logic" which allows this calculation of c-v, but an assumption of Galilean relativity, which we know to be inadequate at higher speeds.

 

[Doc Al]

Yes, you showed that the second firing did 3 times more work, but did you show why 3 times more work didn't add 3 times more velocity to the pellet? No, the only answer you could give was that because K.E. is defined to be 1/2mv^2.

Huh? Given the definition of work (\vec{F}\cdot \vec{s}), and an understanding of Newton's laws, one derives the fact that W = \Delta (1/2mv^2). Yes, we call that quantity "Kinetic Energy", but regardless of the name it is still 1/2mv^2.

If K.E. was undefined and you started from rest and applied a constant force over 1 distance unit to achieve a velocity v for a particular mass, wouldn't it be reasonable to assume that if you applied the same force over 2 distances, you should achieve a final velocity of 2v?

Not if you understand Newton's laws. A force can both do work on a particle leading to a change in KE and exert an impulse (\vec {F}\Delta t) on a particle leading to a change in momentum.

 

[bruce2g]

...You thought the right brick came to rest but it didn't since it has a final kinetic energy of -1.0mv^2.

In my universe, macroscopic kinetic objects cannot have negative energy; therefore I will waste no more time on this matter since you appear to be inhabiting some other universe in which bricks can have negative energy.

Bruce